package com.l.dubbo.service; 
import java.util.HashSet;
import java.util.Set;
import java.util.regex.Matcher;
import java.util.regex.Pattern;
 
public class ThreadPoolTest {
    public static boolean check1(String context, String regEx) {
        Pattern pattern = Pattern.compile(regEx);
        Matcher matcher = pattern.matcher(context);
        String ss = "";
        String s2 = "";
        boolean flag = true;
        int count = 0;
        //符合正則表達(dá)式的子段為特殊符號
        //比較子段是否重復(fù)  若重復(fù)則特殊符號種類不止一種
        while (matcher.find()) {
            if (count == 0) {
                ss = matcher.group();
//                System.out.println(ss);
            } else {
                s2 = matcher.group();
//                System.out.println(s2);
            }
            if (!(ss).equals(s2) && count > 0) {
                flag = false;
            }
            count++;
        }
//        System.out.println(count);
        return flag;
    }
 
    public static boolean check2(String context, String regEx) {
        Pattern pattern = Pattern.compile(regEx);
        Matcher matcher = pattern.matcher(context);
        boolean flag = false;
        //利用set的元素不能重復(fù)，來檢查是否只有一種特殊符號
        Set<String> set = new HashSet<String>();
        while (matcher.find()) {
            set.add(matcher.group());
        }
        if (set.size() == 1) {
            flag = true;
        }
        return flag;
    }
 
    public static void main(String[] args) {
        //s1中只包含'！‘一種特殊字符
        String s1 = "dasd!a22!!!!";
        //s2中包含'！‘和'@‘兩種特殊字符
        String s2 = "dasd!a22@!!!";
        String regEx = "((?=[\\x21-\\x7e]+)[^A-Za-z0-9])";
        System.out.println(check1(s1, regEx));
        System.out.println(check1(s2, regEx));
        System.out.println(check2(s1, regEx));
        System.out.println(check2(s2, regEx));
    }
}

public class Test {
    public static void main(String[] args){
        Pattern pattern = Pattern.compile("^[A-D]+$");
        System.out.println(pattern.matcher("A").find());
        System.out.println(pattern.matcher("E").find());
    }
}